Question

Find $\sin \frac{x}{2},\cos \frac{x}{2},\tan \frac{x}{2}$ in second quadrant, if $\tan x=-\frac{4}{3}$.

Answer 1

Given that x is in second quadrant

$90<x<180$

$45<\frac{x}{2}<90$

$\frac{x}{2}$ lies in first quadrant

So,

$\sin x,\cos x,\tan x$ are positive in first quadrant

$\tan x=\frac{2\tan \left(\frac{x}{2}\right)}{1-{\tan }^2\frac{x}{2}}$

$\frac{-4}{3}=\frac{2\tan \left(\frac{x}{2}\right)}{1-{\tan }^2\frac{x}{2}}$

Solving the above quadratic equation, we get,

$\tan \left(\frac{x}{2}\right)=\frac{-1}{2}$ or $\tan \left(\frac{x}{2}\right)=2$

$\therefore \tan \left(\frac{x}{2}\right)=2(\because \frac{x}{2})$ lies in first quadrant

$1+{\tan }^2\left(\frac{x}{2}\right)={\sec }^2\left(\frac{x}{2}\right)$

$\Rightarrow {\sec }^2\left(\frac{x}{2}\right)=5$

$\Rightarrow \sec \left(\frac{x}{2}\right)=\sqrt{5}$

$\cos \left(\frac{x}{2}\right)=\frac{1}{\sec \left(\frac{x}{2}\right)}$

$\Rightarrow \cos \left(\frac{x}{2}\right)=\frac{1}{\sqrt{5}}$ or $\cos \left(\frac{x}{2}\right)=\frac{\sqrt{5}}{5}$

${\sin }^{2}x+{\cos }^{2}x=1$

$\Rightarrow {\sin }^{2}x=1-{\cos }^{2}x$

$\Rightarrow {\sin }^{2}x=\frac{4}{5}$

$\Rightarrow \sin x=\pm \sqrt{\frac{4}{5}}$